`n_{H_2} = {3,36}/{22,4} = 0,15(mol)`
`2C_nH_{2n+1}OH + 2Na → 2C_nH_{2n+1}ONa + H_2`
`n_{text {ancol}} = 2n_{H_2} =0,15 .2= 0,3(mol)`
`M = {10,3}/{0,3} = 34,33`$(g/mol)$
`M = 14n + 18 = 34,33`$(g/mol)$
`→ n = 1,17`
`→` Hai ancol là `CH_3OH; C_2H_5OH`